To prove \(1+\frac{\cot^2 \alpha}{1+\text{cosec} \alpha}=\text{cosec} \alpha\).

Solution:

Step 1: Use the conversion identity of \(\cot^2 \theta\) into \(\text{cosec}^2 \theta \)

L.H.S \(= 1+\frac{\cot^2 \alpha}{1+\text{cosec} \alpha}\)

As \( \cot^2 \theta = \text{cosec}^2 \theta -1\)

Therefore, L.H.S \(= 1+\frac{\text{cosec}^2 \alpha -1}{1+\text{cosec} \alpha}\)

\(\Rightarrow \) L.H.S \(= 1+\frac{\text{cosec}^2 \alpha -1}{\text{cosec} \alpha+1}\)

Step 2: Use the identity \(a^2-b^2= \left(a+b\right) \left(a-b\right) \) and simplify

\(\Rightarrow \) L.H.S \(= 1+\frac{\left(\text{cosec} \alpha +1\right) \left(\text{cosec} \alpha -1 \right)}{\text{cosec} \alpha+1}\)

\(\Rightarrow \) L.H.S \(= 1+\left(\text{cosec} \alpha -1 \right)\)

\(\Rightarrow \) L.H.S \(= \text{cosec} \alpha =\) R.H.S

Therefore, \(1+\frac{\cot^2 \alpha}{1+\text{cosec} \alpha}=\text{cosec} \alpha\)  proved.